If you were proving uniform continuity on all of ℝ you'd have a problem (it isn't), but here you're proving point-wise continuity, which means you can define your delta in terms of more than just epsilon. Think about why this is for a bit, it was a difficult concept for me to swallow at first.

The right side is equal to 14|a3| + ( don't know how to relate this inequality...)

The way to deal with |a3| is simply that |a3| = |a3|

So if

|x| < 2|a|,
|x2| < 4|a2| ,
|x3| < 8|a3|,
and
|a3| is what it is:​

then

|x3| + |ax2| + |a2x| + |a3| < 8|a3| + 4|a2|∙|a| + 2|a|∙|a2| + |a3|​

This will work fine as long as a ≠ 0.

The case in which a = 0 can be handled pretty easily, after you have conquered the more general case. (It is also true that you can work out a method that handles all values of a, but the algebra will be tougher than that above and that looks as if it has given you enough of a problem.)